Basic Analysis – Gently Done Topological Vector Spaces

6 Basic Analysis
Proposition 1.20 Let (X,T) be a topological space and let K be compact.
Suppose that F is closed and F ⊆ K. Then F is compact. (In other words, closed
subsets of compact sets are compact.)
Proof Let {U α : α ∈ J} be any given open cover of F. We augment this collection
with the open set X \F. This gives an open cover of K;
K ⊆ (X \F)∪ �
α∈J
U α .
Since K is compact, there are elements α 1 ,α 2 ,...,α m in J such that
It follows that
and we conclude that F is compact.
K ⊆ (X \F)∪U α1 ∪U α2 ∪···∪U αm .
F ⊆ U α1 ∪U α2 ∪···∪U αm
We now consider continuity of mappings between topological spaces. The definition
is the obvious rewriting of the standard result from metric space theory.
Definition 1.21 Let (X,T) and (Y,S) be topological spaces and suppose that
f : X → Y is a given mapping. We say that f is continuous if and only if
f −1 (V) ∈ T for any V ∈ S. (By taking complements, this is equivalent to f −1 (F)
being closed for every F closed in Y.)
Many of the standard results concerning continuity in metric spaces have analogues
in this more general setting.
Theorem 1.22 Let (X,T) and (Y,S) be topological spaces with (X,T) compact,
and suppose that f : X → Y is continuous. Then f(X) is compact in (Y,S).
(In other words, the image of a compact space under a continuous mapping is
compact.)
Proof Let {V α } be any given open cover of f(X). Then {f −1 (V α )} is an open
cover of X and so, by compactness, there are indices α 1 ,α 2 ,...,α m such that
It follows that
and hence f(X) is compact.
X = f −1 (V α1 )∪···∪f−1 (V αm ).
f(X) ⊆ V α1 ∪···∪V αm
Department of Mathematics King’s College, London